How should you label a triangle before using either rule?

Capital letters for the angles, and the matching lower-case letter for the side opposite each one. Both rules depend on that pairing, so mislabelling is what makes them produce nonsense.

True or False? Knowing all three sides of a triangle is enough to find every angle.

True. The cosine rule, rearranged, finds any angle from the three side lengths alone. Once one angle is known the sine rule gives the others, so three sides fix the triangle completely.

IGCSEFurther MathsEdexcelFlashcardsGeometry & TrigonometryTrigonometric Formulae & Identities

Trigonometric Formulae & Identities (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontSine & Cosine Formulae
Complete the sine rule for any triangle:
$\frac{a}{\sin A} = \frac{b}{______} = \frac{c}{______}$

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Cards in this collection (22)

Complete the sine rule for any triangle:
$\frac{a}{\sin A} = \frac{b}{______} = \frac{c}{______}$
The completed rule is:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Any two of the three fractions can be used at a time; you never need all three at once.
When can you use the sine rule?
Whenever you have a matching pair, a side together with the angle opposite it, plus one further side or angle.
To find an angle it is easier to use the flipped form $\frac{\sin A}{a} = \frac{\sin B}{b}$ .
Why might the sine rule give you the wrong angle?
Because inverse sine on a calculator always returns an acute angle.
If the diagram shows the angle is obtuse, subtract the calculator's answer from $180 °$ .
How should you label a triangle before using either rule?
Capital letters for the angles, and the matching lower-case letter for the side opposite each one.
Both rules depend on that pairing, so mislabelling is what makes them produce nonsense.
When do you use the cosine rule rather than the sine rule?
When you have two sides and the angle between them, or all three sides, and so no matching pair to work with.
It states $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ , with $A$ the angle sitting between $b$ and $c$ .
How do you rearrange the cosine rule to find an angle?
Make $\cos A$ the subject, giving $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ .
Here $a$ must be the side opposite the angle you want, with $b$ and $c$ the two sides enclosing it.
True or False?
Knowing all three sides of a triangle is enough to find every angle.
True.
The cosine rule, rearranged, finds any angle from the three side lengths alone.
Once one angle is known the sine rule gives the others, so three sides fix the triangle completely.
Complete the formula for the area of any triangle:
$Area =______a b \sin______$
The completed formula is:
$Area = \frac{1}{2} a b \sin C$
The angle $C$ must be the one between the two sides $a$ and $b$ , which is the condition most often missed.
What do the cosine rule and the area formula become when the angle is $90 °$ ?
Since $\cos 90 ° = 0$ , the cosine rule collapses to $a^{2} = b^{2} + c^{2}$ , which is Pythagoras.
Since $\sin 90 ° = 1$ , the area formula becomes $\frac{1}{2} \times base \times height$ .
Define trigonometric identity.
A trigonometric identity is a statement about $\sin$ , $\cos$ or $\tan$ that is true for every value of the angle.
The symbol $\equiv$ is sometimes used in place of an equals sign to stress that it always holds.
Complete the Pythagorean identity:
$\sin^{2} θ +______=______$
The completed identity is:
$\sin^{2} θ + \cos^{2} θ = 1$
Note that $\sin^{2} θ$ means ${(\sin θ)}^{2}$ , the value squared, and not the sine of $θ^{2}$ .
What are the two useful rearrangements of $\sin^{2} θ + \cos^{2} θ = 1$ ?
They are $\sin^{2} θ = 1 - \cos^{2} θ$ and $\cos^{2} θ = 1 - \sin^{2} θ$ .
Each lets you swap one squared function for the other, which is how an equation containing both is reduced to one.
What is $\tan θ$ in terms of $\sin θ$ and $\cos θ$ ?
It is the quotient $\tan θ = \frac{\sin θ}{\cos θ}$ .
This comes straight from the unit circle, where $\cos θ$ and $\sin θ$ are the two coordinates and $\tan θ$ is their ratio.
True or False?
The identity $\sin^{2} θ + \cos^{2} θ = 1$ holds for obtuse and negative angles too.
True.
It holds for every angle, which is exactly what makes it an identity rather than an equation.
On the unit circle it is Pythagoras applied to the point $(\cos θ, \sin θ)$ , and that works whichever quadrant the point is in.
How do you turn $2 \sin^{2} θ - \cos θ = 0$ into an equation in $\cos θ$ only?
Replace $\sin^{2} θ$ with $1 - \cos^{2} θ$ , which gives $2 - 2 \cos^{2} θ - \cos θ = 0$ .
Multiplying through by $- 1$ to make the squared term positive leaves $2 \cos^{2} θ + \cos θ - 2 = 0$ .
How do the plus and minus signs work in the addition formulae?
For $\sin$ the sign on the right matches the one on the left, and for $\cos$ it is the opposite.
For $\tan$ the numerator matches and the denominator is opposite, so $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ .
Complete the double angle formula for sine:
$\sin 2 A =______\sin A______A$
The completed formula is:
$\sin 2 A = 2 \sin A \cos A$
The extra $\cos A$ is what catches people out: doubling the angle is not the same as doubling the function.
What are the three forms of the double angle formula for $\cos 2 A$ ?
They are $\cos 2 A = \cos^{2} A - \sin^{2} A = 1 - 2 \sin^{2} A = 2 \cos^{2} A - 1$ .
The last two come from swapping one squared term using the Pythagorean identity, and you pick whichever leaves the function you want.
What is the double angle formula for $\tan 2 A$ ?
It is $\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$ .
Note the minus sign in the denominator and the $\tan^{2} A$ : it is not simply twice $\tan A$ .
True or False?
The double angle formulae have to be remembered separately from the addition formulae.
False.
Each one is just the addition formula with $B$ set equal to $A$ .
Deriving them that way takes a single line and removes three more formulae from what you have to hold in your head.
How can the addition formulae give the exact value of $\sin 15 °$ ?
Write $15 °$ as $45 ° - 30 °$ and apply the formula for $\sin (A - B)$ .
Every term is then an exact value you already know, so the answer comes out in surds with no calculator needed.
Simplify $\tan (x + \frac{π}{4})$ using the addition formula.
Since $\tan \frac{π}{4} = 1$ , the formula gives $\frac{\tan x + 1}{1 - \tan x}$ .
Spotting that $\tan \frac{π}{4} = 1$ is what makes the expression collapse so neatly.

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